I was having a look at the circuit design of Keithley 236 SMU. You can find this part of the circuit in the Keithley 236 Service Manual.

Circuit Diagram

Q47 is a BJT. Only the base and emitter are connected, which means it is used as a diode.

I also made a simulation to compare these three variations, which seems not a lot of difference. (Pink is voltage, blue is current) Simulation

My questions are:

  • Is it better to use diode-connected BJT to enhance its flowing current without affecting too much the forward voltage in this circuit?
  • Are there any particular reasons against using a normal diode or a diode-connected BJT in the original design?
  • What are the differences among those three kinds of diodes (i.e. diode-connected BJT vs base-emitter BJT diode vs diode)? (I know there is an answer about diode-connected BJT and diode here.)
  • In this design, all these three "diodes" are used. How could I determine which one to use in the design?

My initial guess is the cost and part availability but I would like to know if there is any design consideration about this.

Edit - This is where diode-connected BJTs are used in this design.

additional schematic

  • \$\begingroup\$ What is the circuit, a precision rectifier? It looks to be, but it depends somewhat on the surrounding circuitry. And I have no idea if that makes sense with respect to whatever the input signal is. \$\endgroup\$ May 28, 2022 at 21:09
  • \$\begingroup\$ Yes. There are 4 of this circuit to switch the signal route based on the "Sense" feedback and the set value via DAC. You can see CR10 at the output determines the output polarity. Its output will be fed into an error amplifier to control the final output. \$\endgroup\$
    – ONLYA
    May 28, 2022 at 21:20
  • \$\begingroup\$ Q48 would seem to be another, though actually drawn as a diode this time, although CR10 (presumably "crystal rectifier" which you don't often see nowadays) is an actual diode. Hmmm... Now if Q47,48 were part of the same package, there may be thermal matching going on, to cancel out some voltage drift. \$\endgroup\$
    – user16324
    May 28, 2022 at 21:48
  • \$\begingroup\$ @user_1818839 Q47 and Q48 are not in the same package. They are not even close to each other on the PCB. So I think it is not for thermal matching. \$\endgroup\$
    – ONLYA
    May 28, 2022 at 22:31

3 Answers 3


I suspect, without knowing the specs, that the criteria is for this design is bandwidth of the ideal rectifier (with Op-Amp negative feedback) for the transistor characteristics selected in the Vbe(f) vs Ibe(f).

The Ideality Factor =1 for which the active diode is best for lowest Rce due to current gain in a common size is commonly used for DC current mirrors.

  • schottky diodes might be lower junction capacitance but much higher bulk resistance. CMOS input protection must be faster than then FETs so they are made extremely small and thus only rated for 5mA DC current. This might also be suitable.
  • Silicon diodes might also be a higher RC breakpoint, yet the

But it is the BW of the rectifier for which a transistor is chosen with low Cbe(low pF) in the BE junction , it will perform better in a spectral sweep , carrier rectifier than a small signal diode.

The fact is that diode capacitance increases with size and current ratings at V=0 and bulk resistance Or incremental saturation R also reduces with increased current rating.

Without knowing which diode and transistor characteristics were used, the forward voltage is irrelevant as the Op-Amp loop gain can null the input error.

So my opinion is they selected this for the frequency response of a low cost transistor, rather than a more expensive PIN diode.

This answer might be improved if then application specs were given.

  • \$\begingroup\$ Thanks for your answer! I think this answers my questions. I have not thought about the frequency response before. I will have a look at the spec sheet of this SMU to analyse the spec for this circuit for an improved answer. There are also diode-connected BJTs connected source to the gate of the output MOSFETs. I think it is to protect the MOSFET as well as parallel a capacitor from the BJT between the gate and source. (Please see the updated image in the question) \$\endgroup\$
    – ONLYA
    May 28, 2022 at 23:06
  • \$\begingroup\$ I don't think you'd use a Schottky in that place: the leakage current is way too high \$\endgroup\$
    – LuC
    May 5, 2023 at 12:21
  1. It can be better to use a diode-connected BJT (NPN) as a diode. This is because a BJT usually has significant base resistance. If you just use the B-E junction, the base resistance affects the forward voltage. If you connect B-C, then most of the current flows in the collector, while the forward voltage is determined by the VBE. Since less current flows in the base terminal, base resistance has much less effect. Your simulation shows this.

Be aware that most NPNs have a maximum reverse BVEBO of less than 6 V, so they are not capable of withstanding more than that voltage.

It is not clear what the requirements of the diode-connected Q47 are -- current is quite low, so base R isn't significant. Perhaps it is a device with particularly low capacitance.

  1. Without knowing what the requirements are, it is difficult to understand why a normal diode (e.g. 1N914, or 1N4148) wasn't used.

  2. Leaving the collector open circuit is a strange choice. This means the B-C junction will become forward biased (VCE will be very small, on the order of 10-100 mV) and reverse recovery may become longer than desired. The C-B junction capacitance will become part of the circuit performance.

  • \$\begingroup\$ Thanks for your answer! There are detailed explanations at the system level about how this system will work. The input and feedback are likely within +/-5V. With the high resistance value up to a few kilo-ohms, the current is low. So it sounds like the BJT can withstand the reverse voltage and low current makes the base R not significant. But it still does not make sense why not simply using a diode. \$\endgroup\$
    – ONLYA
    May 28, 2022 at 21:37

The art of asking questions

I am always in awe of such in-depth questions... even more so when they are constructed so well. The art of asking questions is no less valuable than the art of answering them... and that is what makes CE EE so attractive.

I have developed the OP's valuable idea of ​​comparing the three diode circuits by building, exploring and explaining them step by step. To show the evolution of the basic idea, I have inserted additional "shots" in this scenario.

I used to do it through real experiments (see my students story) and now with the help of CircuitLab simulations. To simplify the schematics and make them more conceptual, I have supplied the diode elements with a current source. In real circuits, its role is successfully performed by a humble resistor and voltage source because the voltage drops across the diodes vary little.

My explanations refer to the principles on which circuits built with discrete elements are based, not to technological considerations related to the production of ICs.

How to keep a constant voltage

This is one of the most important diode applications in electronic circuits where diodes stabilize the voltage by shunting; hence the name "shunt regulators". Let's follow the evolution of this powerful idea.

Constant resistor

The simplest way to create a constant voltage V = I.R (Ohm's law) is by passing a constant current I through a constant resistance R.


simulate this circuit – Schematic created using CircuitLab

The problem is when the current is "not so constant" or when a load is connected in parallel to the resistor.

Dynamic resistor

The remedy is very simple and intuitive - make the resistor "self-varying" (dynamic) so that its resistance changes in the opposite direction to the current and V = I.R = const. A very good way to understand this idea is to put ourselves in the place of the resistor and begin to perform its role. Let's do it together with the simulation below.

Imagine that I have reduced the current in half (5 mA) so the voltage across the resistor has dropped to 5 V. However, you sense this and increase the resistance twice (2 k) so the voltage remains 10 V.


simulate this circuit

Or if I have doubled the current (20mA), you halve the resistance (.5k) and the voltage remains 10V.


All types of diodes (ordinary Si and Ge, LED, Zener, Schottky, etc.) behave this way when forward biased. Let's for example investigate the IV curve of an Si diode in the schematic below by the help of the DC sweep simulation. The current through diode changes from 0 to 10 mA and for each step, the voltage across diode is plotted.


simulate this circuit

As you can see in the graph below, in the 0 ÷ 400 mV range, the diode behaves like an ordinary constant resistor with a very high static resistance (R = V/I). After that, however, it decides for some reason to reduce its resistance. As a result, its IV curve becomes almost horizontal and as they say, "its differential resistance tends to zero".

Diode IV curve

The problem is that the diode does have some series resistance R_S (for example, 0.568 ohm for 1N4148), and that changes the slope of the curve. Let's set it to vary as a parameter in the range 0 ÷ 100 ohms with a step of 25 ohm (significantly increased to see the effect).

Diode IV curve R_S par

Base-emitter junction

Interestingly, for better or worse, a transistor base-emitter junction behaves the same way as a diode....


simulate this circuit

... but it seems to have a higher series base resistance B_R (10 ohm for 2N3904).

Base-emitter IV curve

Let's plot a family of base-emitter IV curves at different B_R as above.

Base-emitter IV curve - Rb var

"Active diode"

Basic idea. I don't know who came up with the brilliant idea to simply connect the transistor collector to its base (as later, the op-amp output to its inverting input)... but I am sure it was people familiar with Harold Black's idea of introducing negative feedback in amplifiers. This connection introduces a voltage-type negative feedback that has a drastic effect on the transistor behavior converting it from a current to a voltage "source".

Operation. At the first moment after turning on the current source, the transistor is off. The current flows through the base-emitter junction and the transistor begins to divert current through its collector-emitter part until equilibrium occurs (Ic = beta.Ib); the voltage settles at about 0.7 V. Thus the output collector-emitter part of the transistor is connected in parallel to the input base-emitter junction; so we can think of this combination as a "buffered diode". The true diode (base-emitter junction) diverts only a beta part of the whole input (collector) current; so it acts as a low power (signal) diode that determines the behavior of the power "diode". Most of the current passes through the collector-emitter junction that initially had the behavior of a current stabilizer but now acts as a voltage stabilizer.


simulate this circuit

Conceptual schematic. For the purposes of intuitive understanding, we can imagine the transistor as a "current divider" of two "resistors" in parallel - a low-power dynamic Rbe and a powerful controlled Rce, which interact. I had to play around a bit to adjust their resistances to match the real transistor schematic above. As you can see, Ic/Ib = Rbe/Rce = beta (140 here); the similarity is striking.


simulate this circuit

Now we come to the most interesting part of our study - to see what is the influence of R_B on the IV curve. As we expected, it is insignificant because the base current is "beta" times smaller than the collector current; the IV curves are almost horizontal and the same (as they say, the diode differential resistance in this part is lower). That is why the active diode is better than the ordinary diode. This confirms the assumption of @jp314.

Active diode IV curve - Rb var

Zener diode

Zener diodes have significantly higher voltage and power dissipation than ordinary diodes.


simulate this circuit

Zener diode IV curve

"Transistor Zener diode"

Therefore, it is of interest how to make a "transistor Zener diode". The recipe is simple: If we insert a Zener diode between the collector and the base, we will make the transistor raise its collector voltage with the Zener voltage (Vz is a disturbance for the transistor which it must overcome).


simulate this circuit

'Transistor Zener diode' IV curve

"Rubber Zener diode"

We can achieve almost the same effect with a voltage divider because if we apply not the whole collector-emitter voltage to the base-emitter junction but a part of it, VBE will be multiplied (like in the non-inverting amplifier). The "transistor diode" will act as a "transistor Zener diode" with any desired voltage. This network is widely used as a bias circuit in op-amp and power amplifiers.


simulate this circuit

'Rubber' diode IV curve

The "rubber Zener" voltage can be smoothly adjusted by a potentiometer.


simulate this circuit

How to mirror the current

This the most important active diode application in electronic circuits (mainly in ICs) where a current from one place is "cloned" to another or more places. Furthermore, the direction of the current is reversed.

"Reversed transistor"

The negative feedback has the unique property to reverse the transistor behavior. Usually, the input voltage Vbe controls the output collector current Ic (see STEP 2.2) while here, thanks to the negative feedback, it seems as if the "input" collector current Iin controls the "output" base-emitter voltage Vout. This is because the transistor adjusts its Vbe voltage to pass the collector current supplied to it by the current source; so it acts as a log current-to-voltage converter.


simulate this circuit

"Normal transistor"

Then this "output" voltage can be applied as an input Vbe voltage to a "normal" transistor Q2 to convert it to another current Iout = Iin; so it acts as an antilog voltage-to-current converter.


simulate this circuit

Cascaded transistors

In this way, we obtained a "mirror" copy of the entering input current where the output current has the same magnitude but an opposite direction (again entering instead of exiting). It is interesting that the two cascaded nonlinear converters form a linear converter.


simulate this circuit

Note: I have used a little "trick" to adjust the collector current to exactly 10 mA - I have set a resistance of 860 ohm of the ammeter.


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